# What is the derivative of 1/(x+1) ?

Jun 1, 2015

This function can be rewritten as ${\left(x + 1\right)}^{-} 1$, following the law of exponentials that states ${a}^{-} n = \frac{1}{a} ^ n$.

Now, we can differentiate it using the chain rule, which, in turn, states that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

In this case, if we rename $u = x + 1$, then, we have that the original function $y = {\left(x + 1\right)}^{-} 1$ becomes $y = {u}^{-} 1$ and, now, we can proceed to the chain rule steps:

$\frac{\mathrm{dy}}{\mathrm{du}} = - 1 \cdot {u}^{-} 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

Thus,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- 1 {u}^{-} 2\right) \left(1\right) = - \frac{1}{u} ^ 2 = - \frac{1}{x + 1} ^ 2$